Problem: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 7,\enspace 23,\enspace 14,\enspace 3,\enspace 2$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{7 + 23 + 14 + 3 + 2}{{5}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-2.8$ years $7.84$ years $^2$ $23$ years $13.2$ years $174.24$ years $^2$ $14$ years $4.2$ years $17.64$ years $^2$ $3$ years $-6.8$ years $46.24$ years $^2$ $2$ years $-7.8$ years $60.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{7.84} + {174.24} + {17.64} + {46.24} + {60.84}} {{5}} $ $ {\sigma^2} = \dfrac{{306.8}}{{5}} = {61.36\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{61.36\text{ years}^2}} = {7.8\text{ years}} $ The average gorilla at the zoo is 9.8 years old. There is a standard deviation of 7.8 years.